Our friend Bui posted this question for a month now without any reply so I decided to give a try.
I summit to your consideration an easy visual graph proof which indeed it is not as forceful like a rigorous algebraic proof but as I said I believe is easy to visualize.I start to call the P plane of the paper where our hexagon containing centers of the circles, the tangency point T1 thu T6 and triangle ABC formed by the straight lines T2T3, T4T5 and T6T1.
Now we will consider that the circle C1 thru C6 are the intersection of six spheres which centers we call them by the same names. Some of the centers will be located above; some will be below our plane P but in same vertical lines that our original centers. Obviously the radii of our spheres must be bigger since their intersections with plane P its not equatorial. In another words the center of sphere will be on six of the eight vertexes of an irregular hexahedron.
Now we shall consider three planes
P1 which will contain centers C1C2C3 and line AB since also contain tangency points T2 and T3.
P2 which will contain centers C3C4C5 and line BC since also contain tangency points T4 and T5
P3 which will contain centers C5C5C1 and line CA since also contain tangency points T6 and T1.
Therefore for being intersection between planes
Line BC3 belong simultaneously to planes P1 and P2 intersection
Line CC5 belong simultaneously to planes P2 and P3 intersection
Line AC1 belong simultaneously to planes P3 and P1 intersection
The intersection of these three plane must render a point M which belong the BC3,CC5 and AC1 therefore means the three lines must be concurrent.
I attach a drawing supporting the above unfortunately lines get somewhat mixed but I expect you can follow the argument scheme.
mpdlc